Question

我的正则表达在最好的时候并不好，而且我已经在这几个小时里一直在努力。我想将一个句子解析为多数是单词的部分，但包括带小数和/或引用文本的数字。

我有一个试验台：

import java.util.ArrayList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;


public class Test {

public static void main(String[] args) {
    String Ihave        = "a.X='Foo123. != was here' and "
                        + " T!= v or "
                        + " cat <> dog and "
                        + " x>-15 and "
                        + " \"Peter and Paul\"=\"Mary\" and "
                        + " y< 15.23 bah ";

    String[] Iwant = {"a.X"
                            , "="
                            , "'Foo123. != was here'"
                            , "and" // accidently left off on the first stackoverflow posting 
                            , "T"
                            , "!="
                            , "v"
                            , "or"
                            , "cat"
                            , "<>"
                            , "dog"
                            , "and"
                            , "x"
                            , ">"
                            , "-15"
                            , "and"
                            , "\"Peter and Paul\""
                            , "="
                            , "\"Mary\""
                            , "and"
                            , "y"
                            , "<"
                            , "15.23"
                            , "bah"};

    String quotedtext    = "((\"[^\"]*\"|'[^']*'))";
    String nospaces      = "[^\\s]";
    String alphanumerics = "\\w";

    String trythis = quotedtext + "" 
                    +"|(<>)|(!=)" // group these pairs together
                    +"|("+nospaces+alphanumerics+"*)"
                    +"|(-\\.|[0-9])" // quoted blocks are ok - but the rest are individual characters
                    ;

    Pattern regex = Pattern.compile(trythis);
    Matcher regexMatcher = regex.matcher(Ihave);
    int x=0;
    while (regexMatcher.find()) {
        String parsed = regexMatcher.group();
        if ( x<Iwant.length ) {
            if ( Iwant[x].equals(parsed)) {
                System.out.println(parsed);
            }
            else {
                System.out.println(parsed+"                         but not as expected ("+Iwant[x]+")");
            }
        }
        else {
            System.out.println(parsed+"              but not as expected");
        }
        x++;
    } 

    System.out.println("\ndone");
}

}

当我运行它时，我得到以下内容：

a                         but not as expected (a.X)
.X                         but not as expected (=)
=                         but not as expected ('Foo123. != was here')
'Foo123. != was here'                         but not as expected (and)
and                         but not as expected (T)
T                         but not as expected (!=)
!=                         but not as expected (v)
v                         but not as expected (or)
or                         but not as expected (cat)
cat                         but not as expected (<>)
<>                         but not as expected (dog)
dog                         but not as expected (and)
and                         but not as expected (x)
x                         but not as expected (>)
>                         but not as expected (-15)
-15                         but not as expected (and)
and                         but not as expected ("Peter and Paul")
"Peter and Paul"                         but not as expected (=)
=                         but not as expected ("Mary")
"Mary"                         but not as expected (and)
and                         but not as expected (y)
y                         but not as expected (<)
<                         but not as expected (15.23)
15              but not as expected (bah)
.23              but not as expected
bah              but not as expected

done

虽然我对模式的最后一部分持怀疑态度，但除了全停/小数点（一直被视为单独的单词）之外，一切看起来都很好 - 如何解决这个问题（即如何获得“aX”）和“15.23”待在一起）？

我想我的正则表达式的重点在于不应将点视为分组中断。

任何帮助都将非常感激。谢谢甲

无法使用Java Pattern匹配器解析句子

0 个答案: